ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

By hit and trial,
we put x = 5 and y = 1
so that,
(3x + 7y) = (3 x 5 + 7 x 1) = 22, which is divisible by 11.
(4x + 6y) = ( 4 x 5 + 6 x 1) = 26, which is not divisible by 11;
(x + y + 4 ) = (5 + 1 + 4) = 10, which is not divisible by 11;
(9x + 4y) = (9 x 5 + 4 x 1) = 49, which is not divisible by 11;
(4x - 9y) = (4 x 5 - 9 x 1) = 11, which is divisible by 11.

Since the numbers are co-prime, they contain only 1 as the common factor.

Also, the given two products 551 and 1073 have the middle number in common.

So, middle number = H.C.F. of 551 and 1073 = 29

First number = 551/29 = 19
Third number = 1073/29 = 37

Required sum = (19 + 29 + 37) = 85

Since, p² - n² = 12 and p > n > 0
Best way to solve this question is putting assuming value,
let p = 4 and n = 2
So, 4² - 2² = 12, and the assumption values are correct. So, 4-2 is equal to 2

Let,
The number is x,
therefore, (1/2) x + 4 = 14,
so, x = 20

Let, The numbers are x & y,
therefore,
x - y = 11 ------ (1) and
1/5(x + y) = 9 or, x + y = 45 ------ (2)
adding two equation we got,
2x = 56 or, x = 28,
putting the value of x in equation 1,
we get, y = 17

The odds numbers are x - 4, x - 2, x, x + 2, x + 4 and x + 6
therefore, x + 6 = 15
so, x = 9, hence fourth number is 11

We need to divide each large field into smaller flower beds such that the area of each bed is same.

So, we find the HCF of the larger fields that gives us the area of the smaller field.

HCF (60, 84, 108) = 12

Now, this HCF is the area (in m²) of each flower bed.

Also, area of a rectangular field = Length x Breadth

=> 12 = 6 x Breadth

=> Breadth = 2m

Hence, each flower bed would be 2m wide.

There are two possible values of k 7 and 14.
So, their summation is 21 and 4 will be the remainder when 21 is divided by 17.

a + b + c = 150.

Since, we have to find out the most possible smallest value,
We assume a = 1, b = 2 and c = 147.

So, the median is 2.

x + y = 166,
y + z = 170,
z + 2x = 250.

Solving the equation, y = 84.

First, identify the number that is multiple of 3 more than 100.
That type of number is 102.
So, 102//3 = 34.
Second, we have to identify the number that is multiple of 3 but nearest less than 198.
Now, 198/3 = 66.
Answer is (66 - 34) + 1 = 33

5 / 80

= 0.0625

Find LCM of 2, 3 and 5 is 30. So now divide all options.
Option that will have remainder 0 is the answer.

Let 2nd = 3x,
therefore, 3x + 6x + 2x = 132, x = 12
so, 3x = 36

Let the numbers be a and b. Then, a + b = 12 and ab = 35.
(a + b)/ab = 12/35
arrow 1/b + 1/a = 12/35
Sum of reciprocals of given numbers = 12/35